Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-2x-8y &= 1 \\ 2x+4y &= -2\end{align*}$
Begin by moving the $x$ -term in the second equation to the right side of the equation. $4y = -2x-2$ Divide both sides by $4$ to isolate $y$ $y = {-\dfrac{1}{2}x - \dfrac{1}{2}}$ Substitute this expression for $y$ in the first equation. $-2x-8({-\dfrac{1}{2}x - \dfrac{1}{2}}) = 1$ $-2x + 4x + 4 = 1$ Simplify by combining terms, then solve for $x$ $2x + 4 = 1$ $2x = -3$ $x = -\dfrac{3}{2}$ Substitute $-\dfrac{3}{2}$ for $x$ back into the top equation. $-2( -\dfrac{3}{2})-8y = 1$ $3-8y = 1$ $-8y = -2$ $y = \dfrac{1}{4}$ The solution is $\enspace x = -\dfrac{3}{2}, \enspace y = \dfrac{1}{4}$.